4-force, 4-momentum, energy and mass relations.

[heitor]

This is an exercise of Special Relativity the professor asked last week.
Sorry for the long post, I hope you don't get bored reading it, also, this is my first post here :shy:

Homework Statement

Defining the 4-force that acts on a particle as the proper-time variation of the 4-momentum [itex]F^\mu := \frac{d P^\mu}{d \tau}[/itex].

1. Justify Einsteins relation between mass and energy: [itex] E = mc^2 [/itex]
2. Show, using the [itex]\eta _{\mu\nu}P^\mu P^\nu [/itex], that [itex] E^2 = (mc^2)^2 + (\vec{p}^2c^2) [/itex], where [itex] \vec{p} [/itex] is the 3-momentum.
3. Show that in SR the 3-force and 3-acceleration is not always parallel to each other.

Homework Equations

I'm using the convention: [itex] x^0 = ct [/itex] and [itex] (- + + +) [/itex] for the metric tensor.

4-velocity: [itex] u^\mu = \gamma_{(v)} (c, \vec{v}) [/itex], with [itex] \vec{v} [/itex] the 3-velocity in lab frame.

And some results I got, they are somewhere in Wikipedia and some books also.

gamma: [tex] \frac{d}{d \tau} = \frac{d t}{d \tau} \frac{d}{d t} = \gamma_{(v)} \frac{d}{d t} [/tex]

I'll omit [itex](v)[/itex] in gamma for brevity.

derivative of gamma: [tex] \dot{\gamma} = \frac{d \gamma}{dt} = \gamma ^3 \frac{\vec{v} \cdot \vec{a}}{c^2} [/tex]

acceleration: [tex] a^\mu = \left ( \gamma ^4 \frac{\vec{v} \cdot \vec{a}}{c}, \gamma^4 \vec{a}_\parallel + \gamma ^2 \vec{a}_\perp \right ) [/tex]

where [itex]\vec{a}_\parallel = (\vec{a} \cdot \vec{v}) \vec{v}/v^2[/itex] is the parallel component of the 3-acceleration to the 3-velocity and [itex] \vec{a}_\perp = \vec{a} - \vec{a}_\parallel[/itex] is the perpendicular one.​

3. The Attempt at a Solution for question 1 and 2

I think I got the results, but i have some doubts.

[tex]F^\mu = \frac{d P^\mu}{d \tau} = \left(\frac{d}{d \tau}(m \gamma c), \frac{d}{d \tau}(m \gamma \vec{v}) \right )[/tex]
I'll call this 'result' as (EQ1).

Using the gamma relation in the relevant equations:
[tex]F^\mu = \gamma \left ( m c \dot{\gamma}, \frac{d\vec{p}}{dt} \right ) = \left ( m c \gamma ^4 \frac{\vec{v} \cdot \vec{a}}{c^2}, \gamma \vec{f}_R \right ) [/tex]
I'll call this 'result' as (EQ2). Where [itex] \vec{f}_R [/itex] is the 3-force. We know that [itex] F^\mu [/itex] is mass times acceleration:

[tex]F^\mu = m a^\mu = m \left ( \gamma ^4 \frac{\vec{v} \cdot \vec{a}}{c}, \gamma^4 \vec{a}_\parallel + \gamma ^2 \vec{a}_\perp \right ) [/tex]
This is (EQ3).

Comparing (EQ2) and (EQ3) we get:
[tex] \vec{f}_R = m \left ( \gamma^3 \vec{a}_\parallel + \gamma \vec{a}_\perp \right ) [/tex]

scalar product with \vec{v} gives a variation in energy:
[tex] \vec{f}_R \cdot \vec{v} = m \gamma^3 \vec{a} \cdot \vec{v} = \frac{dE}{d \tau} [/tex]
This is (EQ4).

Here is the derivative with respect to the proper-time, right? I'm not sure about this...​

Putting (EQ4) inside (EQ3) in the first component and re-using the second component of (EQ2):
[tex] F^\mu = \gamma \left (\frac{\vec{f}_R \cdot \vec{v}}{c}, \vec{f}_R \right ) = \left ( \frac{1}{c} \frac{d E}{d\tau}, \frac{d\vec{p}}{d\tau} \right )[/tex]
This is (EQ5).

Comparing (EQ5) and (EQ1) yelds:
[tex]\frac{dE}{d\tau} = \frac{d}{d\tau} (m \gamma c^2)[/tex]

So:

[tex]E = m \gamma c^2 +\ constant[/tex]
This is (EQ6).

This is almost what the question 1 asks, but what is this constant? What is the meaning of [itex]E[/itex]? Is it kinect + 'rest energy'?​

Question 2 I got:

[tex]\eta _{\mu\nu}P^\mu P^\nu = m^2 \gamma ^2 (-c^2 + \vec{v}^2) = -m^2 c^2 [/tex]
(EQ7)

From (EQ6) we can write the 4-momentum as:
[tex] P^\mu = \left ( \frac{E}{c}, \vec{p} \right ) [/tex]
(EQ8)

so:
[tex]\eta _{\mu\nu}P^\mu P^\nu = - \frac{E^2}{c^2} + \vec{p}^2 = -m^2 c^2[/tex]

Wich is the answer to question 2 if the constant of (EQ6) is zero...



4. The attempt at a solution for question 3

If we do the same trick used to split the acceleration in parallel and perpendicular parts:

[tex] \vec{f}_R := \vec{f}_{R, \parallel} + \vec{f}_{R, \perp} [/tex]
We build the parallel one:

[tex] \vec{f}_{R, \parallel} = \frac{(\vec{f}_R \cdot \vec{a}) \vec{a}}{\vec{a}^2} [/tex]
and the perpendicular one:

[tex] \vec{f}_{R, \perp} = \vec{f}_R - \vec{f}_{R, \parallel} = \vec{f}_R - \frac{(\vec{f}_R \cdot \vec{a}) \vec{a}}{\vec{a}^2} [/tex]

But I was unable to prove [itex] \vec{f}_{R, \perp} \neq \vec{0} [/itex].

Do you have any hint? All that I got was some previous equations.


If I was not clear in some statement, please tell me.

(is there a way to put a 'name' in some equations to be displayed on the right side of it? Like latex documents?)
(Sorry for my bad English...)

Thanks in advance,
Heitor.

 

[dauto]

From the expression [tex] \vec{f}_R = m \left ( \gamma^3 \vec{a}_\parallel + \gamma \vec{a}_\perp \right ) [/tex] we see that the force is not parallel to the acceleration because [itex] \gamma^3 [/itex] is different from [itex] \gamma [/itex] so the parallel and perpendicular components scale differently.

 

[dauto]

By the way, I think that instead of taking equations from books and wikipedia, you ought to prove them.

 

[heitor]

From the expression [tex] \vec{f}_R = m \left ( \gamma^3 \vec{a}_\parallel + \gamma \vec{a}_\perp \right ) [/tex] we see that the force is not parallel to the acceleration because [itex] \gamma^3 [/itex] is different from [itex] \gamma [/itex] so the parallel and perpendicular components scale differently.

Is there any interpretation for this? A 3-force in Special Relativity does not have the same meaning as in Newtonian Mechanics?


I did proved them, it was the previous question in the list.

 

[dauto]

The 3-force is derived from force laws just like in Newtonian mechanics but f=ma doesn't apply. instead we have f = m (γ³a_parallel + γa_{perpendicular})